Grades Algebra II & Trigonometry Complex NumbersQuadratics with Complex Roots

Quadratics with Complex Roots

🌀 Complex Numbers

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Quadratics with Complex Roots

Solve quadratic equations whose discriminant is negative.

When the discriminant b² − 4ac is negative, the quadratic equation ax² + bx + c = 0 has no real solutions. Instead it has two complex conjugate solutions. This extends the Fundamental Theorem of Algebra: every degree-n polynomial has exactly n roots in ℂ.

x^2 + 4 = 0 \implies x^2 = -4 \implies x = \pm\sqrt{-4} = \pm 2i

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \quad \text{when } b^2 - 4ac < 0: \sqrt{b^2-4ac} = i\sqrt{4ac-b^2}

Solving with Complex Roots

1Apply the quadratic formula.

2Compute the discriminant Δ = b² − 4ac.

3If Δ < 0, write √Δ = i√|Δ|.

4Simplify to get two complex conjugate solutions.

🔍Solving x² − 2x + 5 = 0

Discriminant: 4 − 20 = −16. Solutions: x = (2 ± √(−16))/2 = (2 ± 4i)/2 = 1 ± 2i. The two roots are 1 + 2i and 1 − 2i.

x = \frac{2 \pm \sqrt{-16}}{2} = \frac{2 \pm 4i}{2} = 1 \pm 2i

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Remember This!

Complex roots of polynomials with real coefficients always come in conjugate pairs a ± bi. If you know one root is 3 − 5i, the other must be 3 + 5i.

✏️ Try It!

What are the solutions to x² + 2x + 10 = 0?

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